一曲线的方程为 $y = \frac{e^{2x}}{x-2}$，其中 $x > 2$。

1. Apply the quotient rule to differentiate the function

Let $u = e^{2x}$ and $v = x-2$.
Then $\dfrac{du}{dx} = 2e^{2x}$ and $\dfrac{dv}{dx} = 1$.
Using the quotient rule $\dfrac{dy}{dx} = \dfrac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$:
$\frac{dy}{dx} = \frac{(x-2)(2e^{2x}) - (e^{2x})(1)}{(x-2)^2}$

2. Simplify the derivative

$\frac{dy}{dx} = \frac{e^{2x}[2(x-2) - 1]}{(x-2)^2}$
$= \frac{e^{2x}(2x - 4 - 1)}{(x-2)^2}$
$= \frac{e^{2x}(2x - 5)}{(x-2)^2}$

1. Set the derivative from part (a) to zero to find the stationary point

From part (a), $\frac{dy}{dx} = \frac{e^{2x}(2x - 5)}{(x-2)^2}$.
For a stationary point, $\frac{dy}{dx} = 0$.
$\frac{e^{2x}(2x - 5)}{(x-2)^2} = 0$

2. Solve for the $x$-coordinate of the stationary point

Since $e^{2x} > 0$ for all $x$, we have:
$2x - 5 = 0$
$x = \frac{5}{2}$

3. Find the corresponding exact $y$-coordinate

Substitute $x = \frac{5}{2}$ into the original equation:
$y = \frac{e^{2(\frac{5}{2})}}{\frac{5}{2} - 2}$
$y = \frac{e^5}{\frac{1}{2}}$
$y = 2e^5$
The coordinates are $(\frac{5}{2}, 2e^5)$.

1. Use the first derivative test to check the sign of the gradient around the stationary point $x = 2.5$.

From earlier parts, $\frac{dy}{dx} = \frac{e^{2x}(2x - 5)}{(x-2)^2}$ and the stationary point is at $x = 2.5$.

For $x$ slightly less than $2.5$ (e.g., $x = 2.4$):
$2x - 5 < 0$, $e^{2x} > 0$, and $(x-2)^2 > 0$, so $\frac{dy}{dx} < 0$.

For $x$ slightly greater than $2.5$ (e.g., $x = 2.6$):
$2x - 5 > 0$, $e^{2x} > 0$, and $(x-2)^2 > 0$, so $\frac{dy}{dx} > 0$.

2. Conclude the nature of the stationary point.

Since the gradient changes from negative to positive as $x$ increases through $2.5$, the stationary point is a minimum point.